H 2 in the minimal basis
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1 H 2 in the minimal basis Alston J. Misquitta Centre for Condensed Matter and Materials Physics Queen Mary, University of London January 27, 2016
2 Overview H 2 : The 1-electron basis. The two-electron basis and the molecular orbitals. Hartree-Fock solution CI solution Desity and density-matrix Dissociation of H 2.
3 1-e basis I The system: Two H-atoms, separated by distance R. We will consider two cases: R = 1.4 Bohr (equilibrium) and R = (dissociation). One electron minimal basis: 1s A (r) = 1 π exp( r A ) 1s B (r) = 1 π exp( r B ) where r A /r A are distances of electron from nucleus A/B.
4 1-e basis II Symmetry-adapted atomic orbitals: where N g = φ 1 (r) = 1σ g = N g [1s A (r) + 1s B (r)] φ 2 (r) = 1σ u = N u [1s A (r) 1s B (r)] 1 2(1 + S) and N u = 1 2(1 S) where S = 1s A (r)1s B (r)dr = (1 + R R2 ) exp( R).
5 1-e basis III All figures from: Molecular Electronic Structure Theory by Helgaker et al..
6 1-e basis IV N-electron basis: 1 Σ + g (g 2 ) = 1σ 2 g = a 1α a 1β vac 1 Σ + g (u 2 ) = 1σ 2 u = a 2α a 2β vac
7 1-e basis V a 2α a 1α vac 3 Σ + u = 1 2 (a 2α a 1β + a 2β a 1α ) vac a 2β a 1β vac 1 Σ + u = 1 2 (a 2α a 1β a 2β a 1α ) vac
8 1-e basis VI What do these term symbols mean? States are classified as gerade (g, German for even ) if they are unchanged by inversion through the centre of mass, or ungerade (u, German for odd ) if inversion changes the sign. The total angular momentum of the molecular state is represented by the central term, in the above case, since we are dealing with orbitals with l = 0 only, we have L = 0 and this is represented by the Σ. The spin multiplicity 2S + 1 of the molecular state is indicated as a superscript on the left. And for Σ states only we may additionally classify states as + if the wavefunction is unchanged on reflection in the plane containing the internuclear axis, or otherwise. The can only arise in open-shell Σ states.
9 Restricted HF I See Szabo & Ostlund secs and What is the matrix element: Ψ H Ψ? Notation: where the core-hamiltonians are: H = h(1) + h(2) + 1 r 12 = O 1 + O 2 h(1) = A Z A r 1A, and similarly for h(2). Here A are all the nuclei.
10 Restricted HF II We will show (in class) that: Ψ O 1 Ψ = 1 h h 2 = h 11 + h 22. and Ψ O 2 Ψ = dx 1 dx 2 χ 1(x 1 )χ 2(x 2 ) 1 χ 1 (x 1 )χ 2 (x 2 ) r 12 dx 1 dx 2 χ 1(x 1 )χ 2(x 2 ) 1 χ 2 (x 1 )χ 1 (x 2 ) r 12 and, defining ij kl = χ i χ j χ k χ l = dx 1 dx 2 χ i (x 1 )χ j (x 2 ) 1 χ k (x 1 )χ l (x 2 ) r 12
11 Restricted HF III we get Ψ O 2 Ψ = The first term is the Coulomb term and the second the exchange term. The two may be written compactly as indicated by the last term. We now write our single determinant energy the Hartree Fock energy as: E HF = Ψ H Ψ = 1 h h
12 Restricted HF IV Now we evaluate the HF energy for 1 Σ + g (g 2 ). Here χ 1 = φ 1 α and χ 2 = φ 1 β. We will integrate out the spin degrees of freedom using: dσα (σ)α(σ) = 1 = dσβ (σ)β(σ) dσα (σ)β(σ) = 0 = dσβ (σ)α(σ) or α β = 0 = β α etc.
13 Restricted HF V 1 h 1 = = = dxχ 1(x)h(r)χ 1 (x) drdσφ 1(r)α (σ)h(r)φ 1 (r)α(σ) dσα (σ)α(σ) drφ 1(r)h(r)φ 1 (r) = 1 (1 h 1) = (1 h 1). Similarly show that = (11 11) g = 0. The second term on the first line is the Chemist s notation used by S&O and the third is the notation used by Helgaker et al.
14 Restricted HF VI So we get the energy of the 1 Σ + g (g 2 ) state as: E(g 2 ) = 2(1 h 1) + (11 11) Q: There is no exchange term present for this state. Why not? Because both spin orbitals in the 1 Σ + g (g 2 ) state have the same spatial part this is referred to as a restricted Hartree Fock (RHF) state. In an unrestricted HF (UHF) state we d allow the up and down spin electrons to reside in different spatial orbitals.
15 Restricted HF VII Summary so far: The single determinant energy the Hartree Fock energy of the ket Ψ = χ 1 χ 2 is: E HF = Ψ H Ψ = 1 h h The energy of the 1 Σ + g (g 2 ) state is: E(g 2 ) = 2(1 h 1) + (11 11)
16 Restricted HF VIII Chemist s Notation ij kl = χ i χ j χ k χ l = dx 1 dx 2 χ i (x 1 )χ j (x 2 ) 1 χ k (x 1 )χ l (x 2 ) r 12 = dx 1 dx 2 χ i (x 1 )χ k (x 1 ) 1 χ j (x 2 )χ l (x 2 ) r 12 = (ik jl) Symmetries are clearer in this notation: (ij kl) = (kl ij) and for real orbitals (the usual case), we additionally have: (ij kl) = (ji kl) = (ij lk) = (ji lk)
17 Restricted HF IX
18 Restricted HF X
19 Restricted HF XI Using the data from table 5.2 we can write down the energies of the H 2 states. In particular, E(g 2 ) = and E(u 2 ) = Hartree. So the bonding state 1 Σ + g (g 2 ) is more strongly bound (compared with two isolated H atoms). Conversely, the anti-bonding state 1 Σ + g (u 2 ) is even more strongly unbound.
20 Density-matrices I The one- and two-electron density We first define the one- and two-electron density matrices: γ 1 (x 1, x 1) = N Ψ (x 1, x 2,, x N )Ψ(x 1, x 2,, x N ) dx 2 dx N γ 2 (x 1, x 2, x 1, x 2) N(N 1) = 2 Ψ (x 1, x 2, x 3,, x N )Ψ(x 1, x 2, x 3,, x N ) dx 3 dx N The density matrices depend on spatial and spin coordinates.
21 Density-matrices II The one-electron and two-electron densities are defined to be the diagonal elements of the density matrices with the spin degrees of freedom integrated out: ρ(r 1 ) = γ 1 (x 1, x 1 )dσ 1 ρ(r 1, r 2 ) = γ 2 (x 1, x 2, x 1, x 2 )dσ 1 dσ 2 Interpretation: The one-electron density ρ(r 1 ) is proportional to the probability of finding an electron at position r 1. The two-electron density ρ(r 1, r 2 ) represents the probability of simultaneously finding two electrons at positions r 1 and r 2 in the molecule.
22 Density-matrices III Let s work out these terms for a 2-e single-det wavefunction: Ψ(x 1, x 2 ) = 2 1/2 (χ 1 (x 1 )χ 2 (x 2 ) χ 2 (x 1 )χ 1 (x 2 )) First evaluate Ψ Ψ: Ψ (x 1, x 2 )Ψ(x 1, x 2) = 1 2 [χ 1(x 1 )χ 2(x 2 )χ 1 (x 1)χ 2 (x 2) + χ 2(x 1 )χ 1(x 2 )χ 2 (x 1)χ 1 (x 2) χ 1(x 1 )χ 2(x 2 )χ 2 (x 1)χ 1 (x 2) χ 2(x 1 )χ 1(x 2 )χ 1 (x 1)χ 2 (x 2)] = 1 2 [χ 1(1)χ 1 (1 )χ 2(2)χ 2 (2 ) + χ 2(1)χ 2 (1 )χ 1(2)χ 1 (2 ) χ 1(1)χ 2 (1 )χ 2(2)χ 1 (2 ) χ 2(1)χ 1 (1 )χ 1(2)χ 2 (2 )]
23 Density-matrices IV Therefore the one-electron density matrix is γ 1 (x 1, x 1) = 2 Ψ (x 1, x 2 )Ψ(x 1, x 2 )dx 2 = χ 1(x 1 )χ 1 (x 1) + χ 2(x 1 )χ 2 (x 1) And using χ i (x) = φ i (r)ω i (σ), the density is ρ(r 1 ) = γ 1 (x 1, x 1 )dσ 1 = φ 1(r 1 )φ 1 (r 1 ) + φ 2(r 1 )φ 2 (r 1 ) In general, for an N-electron single-det wavefunction, ρ(r) = N φ i (r)φ i (r) i=1
24 Density-matrices V The two-electron density matrix is quite simply (no integration needed for the 2-electron wavefunction): γ 2 (x 1, x 2, x 1, x 2) = 2(2 1) Ψ (x 1, x 2 )Ψ(x 2 1, x 2) = 1 2 [χ 1(1)χ 1 (1 )χ 2(2)χ 2 (2 ) + χ 2(1)χ 2 (1 )χ 1(2)χ 1 (2 ) χ 1(1)χ 2 (1 )χ 2(2)χ 1 (2 ) χ 2(1)χ 1 (1 )χ 1(2)χ 2 (2 )] So, if Ψ is a singlet state with χ 1 = φ 1 α and χ 2 = φ 2 β then the two-elecron density is ρ(r 1, r 2 ) = γ 2 (x 1, x 2, x 1, x 2 )dσ 1 dσ 2 = 1 2 [φ 1(1)φ 1 (1)φ 2(2)φ 2 (2) + φ 2(1)φ 2 (1)φ 1(2)φ 1 (2)]
25 RHF:Dissociation I Back to H 2 : For 1σ 2 g = φ 1 α, φ 1 β and 1σ 2 u = φ 2 α, φ 2 β : ρ 1σ 2 g (r) = 2φ 2 1(r) ρ 1σ 2 u (r) = 2φ 2 2(r) ρ 1σ 2 g (r 1, r 2 ) = φ 2 1(r 1 )φ 2 1(r 2 ) ρ 1σ 2 u (r 1, r 2 ) = φ 2 2(r 1 )φ 2 2(r 2 ) Interpretation: Since the two-elecrtron density represents the probability of simultaneously finding two electrons at positions r 1 and r 2 in the molecule, we see here is that the probability of finding an electron at r 1 is unaffected by the electron at r 2. Thus, these single-determinant (Hartree Fock) wavefunctions do not correlate the electrons.
26 RHF:Dissociation II Helgaker et al.
27 RHF:Dissociation III This has consequences for this restricted Hartree Fock (RHF) wavefunction: it does not dissociate into two H-atoms as R. In this limit, S = 1s A (r) 1s B (r) = 0. So φ 1 (r) = 1σ g = 2 1/2 [1s A (r) + 1s B (r)] φ 2 (r) = 1σ u = 2 1/2 [1s A (r) 1s B (r)]. Now let s write 1σ 2 g in terms of the the atomic (non-symmetric) basis functions 1σ 2 g = φ 2 α, φ 2 β = φ 1 (r 1 )φ 1 (r 2 ) 1 2 (α(1)β(2) β(1)α(2))
28 RHF:Dissociation IV Focus on the spatial part φ 1 (r 1 )φ 1 (r 2 ) and use the notation A = 1s A (r) and B = 1s B (r). φ 1 (r 1 )φ 1 (r 2 ) = 1 [A(1)A(2) + B(1)B(2) + A(1)B(2) + B(1)A(2)] 2 = 1 2 A B2 + 1 AB. 2 Here A 2 = A(1)A(2) is the state with both electrons on A, i.e., the state H (similarly for B) and AB = 1 2 [A(1)B(2) + B(1)A(2)] is the state with one electron on A and one on B, i.e. the correctly dissociated state consisting of two neutral H atoms.
29 RHF:Dissociation V Q: Show that the states A 2, B 2 and AB orthonormal. Now consider the RHF energy at dissociation (all cross terms can be shown to tend to vanish as R ): E(g 2 ) = 1σ 2 g H 1σ 2 g = 1 2 A B AB H 1 2 A B AB = 1 4 E(H ) E(H ) (2E(H)) = E(H) E(H ). So, as expected, we do not get 2E(H).
30 RHF:Dissociation VI Q: What happened to the cross terms in expression for E(g 2 )? Show that they all vanish in the R limit. Q: Show the previous result starting from the energy expression for 1σ 2 g : E(g 2 ) = 2(1 h 1) + (11 11) Hint: Expand the symmetry-adapted atomic orbital φ 1 in terms of the 1s A (r) and 1s B (r) basis functions and use E(A 2 ) = E(H ) = 2(A h A) + (AA AA)
31 CI I The Configuration Interaction wavefunction: This is a wavefunction made up of a linear combination of all allowed single determinants. For gerade ground state of H 2 in this minimal basis set this takes the simple form: 1 Σ + g (τ) = cos(τ) 1σ 2 g + sin(τ) 1σ 2 u. No other configurations are allowed to mix as the others are all of ungerade symmetry. There will be two orthogonal solutions, one as above and the other of the form 1 Σ + g (τ + π/2). Q: Show that 1 Σ + g (τ) 1 Σ + g (τ + π/2) = 0.
32 CI II Now we calculate the energy of the system with the CI wavefunction: 1 Σ + g (τ) = cos(τ) 1σ 2 g + sin(τ) 1σ 2 u. The energy of H 2 now becomes (real orbitals): E(τ) = 1 Σ + g (τ) H 1 Σ + g (τ) = cos 2 (τ)e(g 2 ) + sin 2 (τ)e(u 2 ) + 2 sin(τ) cos(τ) 1σ 2 g H 1σ 2 u Q: Show that 1σ2 g H 1σ 2 u = = (12 12). This can also be written as (21 21) = g 2121 due to symmetry of these integrals.
33 CI III So the energy of the CI state is E(τ) = cos 2 (τ)e(g 2 ) + sin 2 (τ)e(u 2 ) + sin(2τ)(21 21). To find the optimum τ we minimize to get tan(2τ) = 2(21 21) E(g 2 ) E(u 2 ) so solutions are τ n = 1 [ ] 2 arctan 2(21 21) E(g 2 ) E(u 2 + nπ ) 2, where n is an integer.
34 CI IV From tables 5.1 and 5.2, for H 2 at its equilibrium separation, we get two solutions: τ 0 = and τ 1 = π/2. Recall that the solutions must be π/2 apart to result in orthogonal states. These give (the e indicates equilibrium separation): 1 Σ + g (τ 0 ) e = σ 2 g σ 2 u 1 Σ + g (τ 1 ) e = σ 2 g σ 2 u I.e., the g.s. is dominated with the HF solution 1σ 2 g with a weight of 98.8%.
35 CI V Here is Table 5.2 from Helgaker et al. again:
36 CI VI Energies of these states are listed in table 5.2. We see that the CI g.s. is 1.4% lower than the 1σ 2 g HF ground state. This may not seem like much, but it is significant. Further, the effect of the CI g.s. on the two-electron density is enormous (fig. 5.5): the small fraction of the 1σ 2 u state introduces what is known as Left Right correlation: the two electrons are now correlated and prefer to sit on opposite nuclei. We will later demonstrate that this mixing of states allows the CI g.s. to correctly dissociate into two H-atoms at R, whereas, the HF g.s. doesn t. The one- and two-electron densities can be calculated as for the RHF wavefunction. These are displayed on the next slide.
37 CI VII Helgaker et al.
38 CI VIII Q: Work out the one- and two-electron densities of the CI wavefunction. As we will show soon, in the R limit, τ 0 = π/4. Write down the two-electron density in this limit and by expressing it in terms of the 1s A (r) and 1s B (r) orbitals, show that the CI wavefunction has indeed introduced Left Right correlation as shown in Fig. 5.5.
39 CI IX Notice the following: There is very little change in the one-electron density from the RHF case. Here the 1σ 2 g (i.e. RHF) state has a weight of 98.8%. The 1σ 2 u state contributing only 1.2%. However, the two-electron density is vastly different. Now it indicates a vanishing probability for the electrons to be on the same atom. Instead, electrons in the CI wavefunction prefer to reside on opposite nuclei. This correlation is called Left Right correlation. It is a non-dynamical correlation that arises when multiple configurations (many-electron determinants) are used to describe the state.
40 CI X Dissociation of the CI wavefunction with an energy 1 Σ + g (τ) = cos(τ) 1σ 2 g + sin(τ) 1σ 2 u. E(τ) = cos 2 (τ)e(g 2 ) + sin 2 (τ)e(u 2 ) + sin(2τ)(12 12) where τ n = 1 [ ] 2 arctan 2(12 12) E(g 2 ) E(u 2 + nπ ) 2
41 CI XI What is τ in the dissociation limit? For R we have (show it!): E(g 2 ) = E(u 2 ) = 2h AA (AA AA) This degeneracy can be expected on physical grounds. Also, in the R limit (12 12) = 1 ((A(1) + B(1))(A(1) B(1)) (A(2) + B(2))(A(2) B(2))) 4 = 1 [(AA AA) + (BB BB)] 4 = 1 2 (AA AA) 0 Here we have used the fact that any cross-terms involving A and B will vanish in the large-r limit.
42 CI XII Consequently, for R, 2(21 21), so E(g 2 ) E(u 2 ) τ n = π 4 + nπ 2. The ground state is n = 0, or τ 0 = π 4 and we get 1 Σ + g (τ) 1 2 [ 1σ 2 g 1σ 2 u ] E(τ) 1 2 (E(g 2 ) + E(u 2 )) (12 12)
43 CI XIII Q: Show that in this limit 1 Σ + g (τ) correctly describes two H- atoms. I.e., show that 1 Σ + g (τ) = AB. Using the results we have stated (and you have to prove) earlier, we get E( π/4) = 1 2 (E(g 2 ) + E(u 2 )) (12 12) = 2h AA (AA AA) 1 2 (AA AA) = 2h A A = 2E(H). I.e., the CI energy correctly tends to the energy of 2 hydrogen atoms as R.
44 CI XIV In summary: H 2 (RHF) H H H 2 (FCI) 2H
45 UHF I CI is computationally expensive. In general there are a lot of determinants possible and the variational space increases exponentially with the size of the basis. So it would be nice to have an alternative way to dissociate H 2. There is one: the unrestricted Hartree Fock (UHF) method. Here we realise that at dissociation we want the spatial parts of orbitals used by the two electrons to be different: the α-spin electron will be associated with one hydrogen atom and the β-spin electron with the other. So we need to allow our single determinant this freedom. This leads to the UHF solution.
46 UHF II Define the UHF wavefunction Ψ = χ α 1 χβ 1 where the unrestricted spin-orbitals are defined to be where χ α 1 (x) = ψ 1α (r)α(ω) χ β 1 (x) = ψ 1β(r)β(ω) ψ 1α = cos(θ)φ 1 + sin(θ)φ 2 ψ 1β = cos(θ)φ 1 sin(θ)φ 2 Q: Show that this choice for the spatial orbitals covers all possibilities. I.e., that for θ = 0 we get the RHF solution and for θ = π/4 we get the dissociated limit of 2 H-atoms.
47 UHF III Rather than solve the UHF problem for you, I will outline it and expect you to solve it completely for homework. This is an important problem so I require you to write it up and submit it to me! Next write down the energy of this UHF wavefunction. Start from the general form for the energy of a single determinant state (we proved this at the start of this lecture): E = Ψ H Ψ = 1 h h Show that the last term vanishes. Write each of the terms in the energy expression in terms of g = φ 1 and u = φ 2. I will use g and u are short forms for these orbitals in the expressions below.
48 UHF IV Hence show that the energy can be written as a function of the angle θ: E(θ) = 2 cos 2 (θ)h gg + 2 sin 2 (θ)h uu + 2 cos 4 (θ)(gg gg) + 2 sin 4 (θ)(uu uu) + 2 sin 2 (θ) cos 2 (θ)[(gg uu) 2(gu gu)]. Find the extrema of E(θ). There should be two solutions. Characterize the solutions: they are not both minima so you will need to find the second derivative of E(θ). Do this carefully. Use integral values from table 5.1 to make a plot of the energy as a function of θ at R e and at dissociation (R ). Do your results agree with this plot? (use any plotting package - but Mathematica or Gnuplot may be best suited for this)
49 UHF V For the θ 0 solution: evaluate all matrix elements in the R limit and show that in this limit θ π/4. Hence show that the UHF energy in this limit is that of two H-atoms. Solve this correctly and completely and you will have understood everything we have covered so far.
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